# Pde Chapter1

Chapter 1 Heat Equation 1. 1 Introduction We wish to discuss the solution of elementary problems involving partial di? erential equations, the kinds of problems that arise in various ? elds of science and engineering. A partial di? erential equation (PDE) is a mathematical equation containing partial derivatives, for example, ? u ? u +3 = 0. ?t ? x (1. 1. 1) We could begin our study by determining what functions u(x, t) satisfy (1. 1. 1). However, we prefer to start by investigating a physical problem. We do this for two reasons.

First, our mathematical techniques probably will be of greater interest to you when it becomes clear that these methods analyze physical problems. Second, we will actually ? nd that physical considerations will motivate many of our mathematical developments. Many diverse subject areas in engineering and the physical sciences are dominated by the study of partial di? erential equations. No list could be all-inclusive. However, the following examples should give you a feeling for the type of areas that are highly dependent on the study of partial di? erential equations: acoustics, aerodynamics, elasticity, electrodynamics, ? id dynamics, geophysics (seismic wave propagation), heat transfer, meteorology, oceanography, optics, petroleum engineering, plasma physics (ionized liquids and gases), and quantum mechanics. We will follow a certain philosophy of applied mathematics in which the analysis of a problem will have three stages: 1. Formulation 2. Solution 3. Interpretation We begin by formulating the equations of heat ? ow describing the transfer of thermal energy. Heat energy is caused by the agitation of molecular matter. Two 1 2 Chapter 1. Heat Equation basic processes take place in order for thermal energy to move: conduction nd convection. Conduction results from the collisions of neighboring molecules in which the kinetic energy of vibration of one molecule is transferred to its nearest neighbor.Thermal energy is thus spread by conduction even if the molecules themselves do not move their location appreciably. In addition, if a vibrating molecule moves from one region to another, it takes its thermal energy with it. This type of movement of thermal energy is called convection. In order to begin our study with relatively simple problems, we will study heat ? ow only in cases in which the conduction of heat energy is much more signi? ant than its convection. We will thus think of heat ? ow primarily in the case of solids, although heat transfer in ? uids (liquids and gases) is also primarily by conduction if the ? uid velocity is su? ciently small. 1. 2 Derivation of the Conduction of Heat in a One-Dimensional Rod Thermal energy density. We begin by considering a rod of constant crosssectional area A oriented in the x-direction (from x = 0 to x = L) as illustrated in Fig. 1. 2. 1. We temporarily introduce the amount of thermal energy per unit volume as an unknown variable and call it the thermal energy density: e(x, t) ? hermal energy density. We assume that all thermal quantities are constant across a section; the rod is onedimensional. The simplest way this may be accomplished is to insulate perfectly the lateral surface area of the rod. Then no thermal energy can pass through the lateral surface. The dependence on x and t corresponds to a situation in which the rod is not uniformly heated; the thermal energy density varies from one cross section to another. z ?(x,t) A ? (x + ? x,t) y x x=0 x x + ? x x=L Figure 1. 2. 1 One-dimensional rod with heat energy ? owing into and out of a thin slice. Heat energy.

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We consider a thin slice of the rod contained between x and x+ ? x as illustrated in Fig. 1. 2. 1. If the thermal energy density is constant throughout the volume, then the total energy in the slice is the product of the thermal energy 1. 2 Conduction of Heat in One-Dimension 3 density and the volume. In general, the energy density is not constant. However, if ? x is exceedingly small, then e(x, t) may be approximated as a constant throughout the volume so that heat energy = e(x, t)A ? x, since the volume of a slice is A ? x. Conservation of heat energy. The heat energy between x and x + ? changes in time due only to heat energy ? owing across the edges (x and x+? x) and heat energy generated inside (due to positive or negative sources of heat energy). No heat energy changes are due to ? ow across the lateral surface, since we have assumed that the lateral surface is insulated. The fundamental heat ? ow process is described by the word equation rate of change of heat energy in time heat energy ? owing across boundaries per unit time = + heat energy generated inside per unit time. This is called conservation of heat energy. For the small slice, the rate of change of heat energy is ? [e(x, t)A ? x] , ? where the partial derivative ? ?t is used because x is being held ? xed. Heat ? ux. Thermal energy ? ows to the right or left in a one-dimensional rod. We introduce the heat ? ux: ? (x, t) = heat ? ux (the amount of thermal energy per unit time ? owing to the right per unit surface area). If ? (x, t) < 0, it means that heat energy is ? owing to the left. Heat energy ? owing per unit time across the boundaries of the slice is ? (x, t)A ? ?(x + ? x, t)A, since the heat ? ux is the ? ow per unit surface area and it must be multiplied by the surface area. If ? (x, t) > 0 and ? (x + ? x, t) > 0, as illustrated in Fig. . 2. 1, then the heat energy ? owing per unit time at x contributes to an increase of the heat energy in the slice, whereas the heat ? ow at x + ? x decreases the heat energy. Heat sources. We also allow for internal sources of thermal energy: Q(x, t) = heat energy per unit volume generated per unit time, 4 Chapter 1. Heat Equation perhaps due to chemical reactions or electrical heating. Q(x, t) is approximately constant in space for a thin slice, and thus the total thermal energy generated per unit time in the thin slice is approximately Q(x, t)A ? x. Conservation of heat energy (thin slice).

The rate of change of heat energy is due to thermal energy ? owing across the boundaries and internal sources: ? [e(x, t)A ? x] ? ?(x, t)A ? ?(x + ? x, t)A + Q(x, t)A ? x. (1. 2. 1) ? t Equation (1. 2. 1) is not precise because various quantities were assumed approximately constant for the small cross-sectional slice. We claim that (1. 2. 1) becomes increasingly accurate as ? x > 0. Before giving a careful (and mathematically rigorous) derivation, we will just attempt to explain the basic ideas of the limit process, ? x > 0. In the limit as ? x > 0, (1. 2. 1) gives no interesting information, namely, 0 = 0. However, if we ? st divide by ? x and then take the limit as ? x > 0, we obtain ? (x, t) ? ?(x + ? x, t) ? e = lim + Q(x, t), (1. 2. 2) ? x>0 ? t ? x where the constant cross-sectional area has been cancelled. We claim that this result is exact (with no small errors), and hence we replace the ? in (1. 2. 1) by = in (1. 2. 2). In this limiting process, ? x > 0, t is being held ? xed. Consequently, from the de? nition of a partial derivative, ?? ?e =? + Q. ?t ? x (1. 2. 3) Conservation of heat energy (exact). An alternative derivation of conservation of heat energy has the advantage of our not being restricted to small slices.

The resulting approximate calculation of the limiting process (? x > 0) is avoided. We consider any ? nite segment (from x = a to x = b) of the original onedimensional rod (see Fig. 1. 2. 2). We will investigate the conservation of heat energy b in this region. The total heat energy is a e(x, t)A dx, the sum of the contributions of the in? nitesimal slices. Again it changes only due to heat energy ? owing through the side edges (x = a and x = b) and heat energy generated inside the region, and thus (after canceling the constant A) d dt b b e dx = ? (a, t) ? ?(b, t) + a a Q dx. 1. 2. 4) Technically, an ordinary derivative d/dt appears in (1. 2. 4) since only on t, not also on x. However, d dt b b b a e dx depends e dx = a a ?e dx, ? t 1. 2 Conduction of Heat in One-Dimension 5 ?(a,t) ?(b,t) 0 x=a x=b L Figure 1. 2. 2 Heat energy ? owing into and out of a ? nite segment of a rod. if a and b are constants (and if e is continuous). This holds since inside the integral the ordinary derivative now is taken keeping x ? xed, and hence it must be replaced by a partial derivative. Every term in (1. 2. 4) is now an ordinary integral if we notice that b ?? x ? (a, t) ? ?(b, t) = ? a ? x (this1 being valid if ? is continuously di? erentiable). Consequently, b a ?e ?? + ? Q dx = 0. ?t ? x This integral must be zero for arbitrary a and b; the area under the curve must be zero for arbitrary limits. This is possible only if the integrand itself is identically zero. 2 Thus, we rederive (1. 2. 3) as ?? ?e =? + Q. ?t ? x (1. 2. 5) Equation (1. 2. 4), the integral conservation law, is more fundamental than the di? erential form (1. 2. 5). Equation (1. 2. 5) is valid in the usual case in which the physical variables are continuous.

A further explanation of the minus sign preceding ?? /? x is in order. For example, if ?? /? x > 0 for a ? x ? b, then the heat ? ux ? is an increasing function of x. The heat is ? owing greater to the right at x = b than at x = a (assuming that b > a). Thus (neglecting any e? ects of sources Q), the heat energy must decrease between x = a and x = b, resulting in the minus sign in (1. 2. 5). is one of the fundamental theorems of calculus. b proofs of this result are inelegant. Suppose that f (x) is continuous and a f (x) dx = 0 for arbitrary a and b. We wish to prove f (x) = 0 for all x.

We can prove this by assuming that there exists a point x0 such that f (x0 ) = 0 and demonstrating a contradiction. If f (x0 ) = 0 and f (x) is continuous, then there exists some region near x0 in which f (x) is of one sign. Pick a b and b to be in this region, and hence a f (x)dx = 0 since f (x) is of one sign throughout. This 2 Most 1 This contradicts the statement that (1. 2. 5) follows. b a f (x)dx = 0, and hence it is impossible for f (x0 ) = 0. Equation 6 Chapter 1. Heat Equation Temperature and speci? c heat. We usually describe materials by their temperature, u(x, t) = temperature, not their thermal energy density.

Distinguishing between the concepts of temperature and thermal energy is not necessarily a trivial task. Only in the mid-1700s did the existence of accurate experimental apparatus enable physicists to recognize that it may take di? erent amounts of thermal energy to raise two di? erent materials from one temperature to another larger temperature. This necessitates the introduction of the speci? c heat (or heat capacity): speci? c heat (the heat energy that must be supplied to a unit mass of a substance to raise its temperature one unit). c = In general, from experiments (and our de? ition) the speci? c heat c of a material depends on the temperature u. For example, the thermal energy necessary to raise a unit mass from 0? C to 1? C could be di? erent from that needed to raise the mass from 85? C to 86? C for the same substance. Heat ? ow problems with the speci? c heat depending on the temperature are mathematically quite complicated. (Exercise 1. 2. 6 brie? y discusses this situation. ) Often for restricted temperature intervals, the speci? c heat is approximately independent of the temperature. However, experiments suggest that di? erent materials require di? rent amounts of thermal energy to heat up. Since we would like to formulate the correct equation in situations in which the composition of our one-dimensional rod might vary from position to position, the speci? c heat will depend on x, c = c(x). In many problems the rod is made of one material (a uniform rod), in which case we will let the speci? c heat c be a constant. In fact, most of the solved problems in this text (as well as other books) correspond to this approximation, c constant. Thermal energy. The thermal energy in a thin slice is e(x, t)A ? x. However, it is also de? ed as the energy it takes to raise the temperature from a reference temperature 0? to its actual temperature u(x, t). Since the speci? c heat is independent of temperature, the heat energy per unit mass is just c(x)u(x, t). We thus need to introduce the mass density ? (x): ? (x) = mass density (mass per unit volume), allowing it to vary with x, possibly due to the rod being composed of nonuniform material. The total mass of the thin slice is ? A ? x. The total thermal energy in any thin slice is thus c(x)u(x, t) · ? A ? x, so that e(x, t)A ? x = c(x)u(x, t)? A ? x. 1. 2

Conduction of Heat in One-Dimension In this way we have explained the basic relationship between thermal energy and temperature: e(x, t) = c(x)? (x)u(x, t). (1. 2. 6) This states that the thermal energy per unit volume equals the thermal energy per unit mass per unit degree times the temperature times the mass density (mass per unit volume). When the thermal energy density is eliminated using (1. 2. 6), conservation of thermal energy, (1. 2. 3) or (1. 2. 5), becomes ?? ?u =? + Q. ?t ? x c(x)? (x) (1. 2. 7) Fourier’s law. Usually, (1. 2. 7) is regarded as one equation in two unknowns, the temperature u(x, t) and the heat ? ux (? w per unit surface area per unit time) ? (x, t). How and why does heat energy ? ow? In other words, we need an expression for the dependence of the ? ow of heat energy on the temperature ? eld. First we summarize certain qualitative properties of heat ? ow with which we are all familiar: 1. If the temperature is constant in a region, no heat energy ? ows 2. If there are temperature di? erences, the heat energy ? ows from the hotter region to the colder region. 3. The greater the temperature di? erences (for the same material), the greater is the ? ow of heat energy. 4. The ? ow of heat energy will vary for di? rent materials, even with the same temperature di? erences. Fourier (1768–1830) recognized properties 1 through 4 and summarized them (as well as numerous experiments) by the formula ? u , ? x ? = ? K0 (1. 2. 8) known as Fourier’s law of heat conduction. Here ? u/? x is the derivative of the temperature; it is the slope of the temperature (as a function of x for ? xed t); it represents temperature di? erences (per unit length). Equation (1. 2. 8) states that the heat ? ux is proportional to the temperature di? erence (per unit length). If the temperature u increases as x increases (i. e. , the temperature is hotter to the right), ? /? x > 0, then we know (property 2) that heat energy ? ows to the left. This explains the minus sign in (1. 2. 8). We designate the coe? cient of proportionality K0 . It measures the ability of the material to conduct heat and is called the thermal conductivity. Experiments 8 Chapter 1. Heat Equation indicate that di? erent materials conduct heat di? erently; K0 depends on the particular material. The larger K0 is, the greater the ? ow of heat energy with the same temperature di? erences. A material with a low value of K0 would be a poor conductor of heat energy (and ideally suited for home insulation).

For a rod composed of di? erent materials, K0 will be a function of x. Furthermore, experiments show that the ability to conduct heat for most materials is di? erent at di? erent temperatures, K0 (x, u) . However, just as with the speci? c heat c, the dependence on the temperature is often not important in particular problems. Thus, throughout this text we will assume that the thermal conductivity K0 only depends on x, K0 (x). Usually, in fact, we will discuss uniform rods in which K0 is a constant. Heat equation. If Fourier’s law, (1. 2. 8), is substituted into the conservation of heat energy equation, (1. 2. 7), a partial di? rential equation results: ? ?u = ? t ? x ? u ? x c? K0 + Q. (1. 2. 9) We usually think of the sources of heat energy Q as being given, and the only unknown being the temperature u(x, t). The thermal coe? cients c, ? , K0 all depend on the material and hence may be functions of x. In the special case of a uniform rod, in which c, ? , K0 are all constants, the partial di? erential equation (1. 2. 9) becomes c? ?u ? 2u = K0 2 + Q. ?t ? x If, in addition, there are no sources, Q = 0, then after dividing by the constant c? , the partial di? erential equation becomes ? 2u ? u = k 2, ? t ? x where the constant k, k= K0 c? 1. 2. 10) is called the thermal di? usivity, the thermal conductivity divided by the product of the speci? c heat and mass density. Equation (1. 2. 10) is often called the heat equation; it corresponds to no sources and constant thermal properties. If heat energy is initially concentrated in one place, (1. 2. 10) will describe how the heat energy spreads out, a physical process known as di? usion. Other physical quantities besides temperature smooth out in much the same manner, satisfying the same partial di? erential equation (1. 2. 10). For this reason (1. 2. 10) is also known as the di? usion equation.

For example, the concentration u(x, t) of chemicals (such as perfumes and pollutants) satis? es the di? usion equation (1. 2. 8) in certain one-dimensional situations. 1. 2 Conduction of Heat in One-Dimension 9 Initial conditions. The partial di? erential equations describing the ? ow of heat energy, (1. 2. 9) or (1. 2. 10), have one time derivative. When an ordinary di? erential equation has one derivative, the initial value problem consists of solving the di? erential equation with one initial condition. Newton’s law of motion for the position x of a particle yields a second-order ordinary di? erential equation, md2 x/dt2 = forces.

It involves second derivatives. The initial value problem consists of solving the di? erential equation with two initial conditions, the initial position x and the initial velocity dx/dt. From these pieces of information (including the knowledge of the forces), by solving the di? erential equation with the initial conditions, we can predict the future motion of a particle in the x-direction. We wish to do the same process for our partial di? erential equation, that is, predict the future temperature. Since the heat equations have one time derivative, we must be given one initial condition (IC) (usually at t = 0), the initial temperature.

It is possible that the initial temperature is not constant, but depends on x. Thus, we must be given the initial temperature distribution, u(x, 0) = f (x). Is this enough information to predict the future temperature? We know the initial temperature distribution and that the temperature changes according to the partial di? erential equation (1. 2. 9) or (1. 2. 10). However, we need to know that happens at the two boundaries, x = 0 and x = L. Without knowing this information, we cannot predict the future. Two conditions are needed corresponding to the second spatial derivatives present in (1. 2. 9) or (1. 2. 0), usually one condition at each end. We discuss these boundary conditions in the next section. Di? usion of a chemical pollutant. Let u(x, t) be the density or concentration of the chemical per unit volume. Consider a one dimensional region (Fig. 1. 2. 2) between x = a and x = b with constant cross-sectional area A. The b total amount of the chemical in the region is a u(x, t)A dx. We introduce the ? ux ? (x, t) of the chemical, the amount of the chemical per unit surface ? owing to the right per unit time. The rate of change with respect to time of the total amount of chemical in the region equals the amount of chemical owing in per unit time minus the amount of chemical ? owing out per unit time.

Thus, after canceling the constant cross-sectional area A, we obtain the integral conservation law for the chemical concentration: d dt Since b ? u ( a ? t d dt b u(x, t) dx a ?? ?x ) dx = 0. b u(x, t) dx = ? (a, t) ? ?(b, t). a (1. 2. 11) = b ? u a ? t dx and ? (a, t) ? ?(b, t) = ? b ?? a ? x dx, it follows that + Since the integral is zero for arbitrary regions, the integrand must be zero, and in this way we derive the di? erential conservation law for the chemical concentration: 10 Chapter 1. Heat Equation ?u ?? + = 0. ?t ? x 1. 2. 12) In solids, chemicals spread out from regions of high concentration to regions of low concentration. According to Fick’s law of di? usion, the ? ux is proportional to ? u ? x the spatial derivative of the chemical concentration: ? = ? k ? u . ?x (1. 2. 13) If the concentration u(x, t) is constant in space, there is no ? ow of the chemical. If the chemical concentration is increasing to the right ( ? u > 0), then atoms of ? x chemicals migrate to the left, and vice versa. The proportionality constant k is called the chemical di? usivity, and it can be measured experimentally. When Fick’s law (1. . 13) is used in the basic conservation law (1. 2. 12), we see that the chemical concentration satis? es the di? usion equation: ? 2u ? u = k 2, ? t ? x (1. 2. 14) since we are assuming as an approximation that the di? usivity is constant. Fick’s law of di? usion for chemical concentration is analogous to Fourier’s law for heat di? usion. Our derivations are quite similar. EXERCISES 1. 2 1. 2. 1. Brie? y explain the minus sign: (a) in conservation law (1. 2. 3) or (1. 2. 5) if Q = 0 (b) in Fourier’s law (1. 2. 8) (c) in conservation law (1. 2. 12), ? u = ? ?? ?t ? x (d) in Fick’s law (1. 2. 13) 1. 2. 2.

Derive the heat equation for a rod assuming constant thermal properties and no sources. (a) Consider the total thermal energy between x and x + ? x. (b) Consider the total thermal energy between x = a and x = b. 1. 2. 3. Derive the heat equation for a rod assuming constant thermal properties with variable cross-sectional area A(x) assuming no sources by considering the total thermal energy between x = a and x = b. 1. 2 Conduction of Heat in One-Dimension 1. 2. 4. Derive the di? usion equation for a chemical pollutant. 11 (a) Consider the total amount of the chemical in a thin region between x and x + ? . (b) Consider the total amount of the chemical between x = a and x = b. 1. 2. 5. Derive an equation for the concentration u(x, t) of a chemical pollutant if the chemical is produced due to chemical reaction at the rate of ? u(? ? u) per unit volume. Suppose that the speci? c heat is a function of position and temperature, c(x, u). (a) Show that the heat energy per unit mass necessary to raise the temperature of a thin slice of thickness ? x from 0? to u(x, t) is not c(x)u(x, t), u ? u but instead 0 c(x, u) d?. (b) Rederive the heat equation in this case. Show that (1. 2. 3) remains unchanged. 1. . 7. Consider conservation of thermal energy (1. 2. 4) for any segment of a onedimensional rod a ? x ? b. By using the fundamental theorem of calculus, ? ?b b 1. 2. 6. f (x) dx = f (b), a derive the heat equation (1. 2. 9). *1. 2. 8. 1. 2. 9. If u(x, t) is known, give an expression for the total thermal energy contained in a rod (0 < x < L). Consider a thin one-dimensional rod without sources of thermal energy whose lateral surface area is not insulated. (a) Assume that the heat energy ? owing out of the lateral sides per unit surface area per unit time is w(x, t). Derive the partial di? rential equation for the temperature u(x, t). (b) Assume that w(x, t) is proportional to the temperature di? erence between the rod u(x, t) and a known outside temperature ? (x, t). Derive that c? ? ? u = ? t ? x K0 ? u ? x ? P [u(x, t) ? ?(x, t)]h(x), A (1. 2. 15) where h(x) is a positive x-dependent proportionality, P is the lateral perimeter, and A is the cross-sectional area. (c) Compare (1. 2. 15) to the equation for a one-dimensional rod whose lateral surfaces are insulated, but with heat sources. (d) Specialize (1. 2. 15) to a rod of circular cross section with constant thermal properties and 0? outside temperature. 2 Chapter 1. Heat Equation *(e) Consider the assumptions in part (d). Suppose that the temperature in the rod is uniform [i. e. , u(x, t) = u(t)].

Determine u(t) if initially u(0) = u0 . 1. 3 Boundary Conditions In solving the heat equation, either (1. 2. 9) or (1. 2. 10), one boundary condition (BC) is needed at each end of the rod. The appropriate condition depends on the physical mechanism in e? ect at each end. Often the condition at the boundary depends on both the material inside and outside the rod. To avoid a more di? cult mathematical problem, we will assume that the outside environment is known, not signi? antly altered by the rod. Prescribed temperature. In certain situations, the temperature of the end of the rod, for example, x = 0, may be approximated by a prescribed temperature, (1. 3. 1) u(0, t) = uB (t), where uB (t) is the temperature of a ? uid bath (or reservoir) with which the rod is in contact. Insulated boundary. In other situations it is possible to prescribe the heat ? ow rather than the temperature, ? K0 (0) ? u (0, t) = ? (t), ? x (1. 3. 2) where ? (t) is given. This is equivalent to giving one condition for the ? rst derivative, ? u/? x, at x = 0. The slope is given at x = 0. Equation (1. 3. ) cannot be integrated in x because the slope is known only at one value of x. The simplest example of the prescribed heat ? ow boundary condition is when an end is perfectly insulated (sometimes we omit the “perfectly”). In this case there is no heat ? ow at the boundary. If x = 0 is insulated, then ? u (0, t) = 0. ?x (1. 3. 3) Newton’s law of cooling. When a one-dimensional rod is in contact at the boundary with a moving ? uid (e. g. , air), then neither the prescribed temperature nor the prescribed heat ? ow may be appropriate. For example, let us imagine a very warm rod in contact with cooler moving air.

Heat will leave the rod, heating up the air. The air will then carry the heat away. This process of heat transfer is called convection. However, the air will be hotter near the rod. Again, this is a complicated problem; the air temperature will actually vary with distance from the rod (ranging between the bath and rod temperatures). Experiments show that, as a good approximation, the heat ? ow leaving the rod is proportional to the temperature 1. 3. Boundary Conditions 13 di? erence between the bar and the prescribed external temperature. This boundary condition is called Newton’s law of cooling.

If it is valid at x = 0, then ? K0 (0) ? u (0, t) = ? H[u(0, t) ? uB (t)], ? x (1. 3. 4) where the proportionality constant H is called the heat transfer coe? cient (or the convection coe? cient). This boundary condition3 involves a linear combination of u and ? u/? x. We must be careful with the sign of proportionality. If the rod is hotter than the bath [u(0, t) > uB (t)], then usually heat ? ows out of the rod at x = 0. Thus, heat is ? owing to the left, and in this case the heat ? ow would be negative. That is why we introduced a minus sign in (1. 3. 4) (with H > 0).

The same conclusion would have been reached had we assumed that u(0, t) < uB (t). Another way to understand the signs in (1. 3. 4) is to again assume that u(0, t) > uB (t). The temperature is hotter to the right at x = 0, and we should expect the temperature to continue to increase to the right. Thus, ? u/? x should be positive at x = 0. Equation (1. 3. 4) is consistent with this argument. In Exercise 1. 3. 1 you are asked to derive, in the same manner, that the equation for Newton’s law of cooling at a right end point x = L is ? K0 (L) ? u (L, t) = H[u(L, t) ? uB (t)], ? x (1. 3. 5) here uB (t) is the external temperature at x = L. We immediately note the signi? cant sign di? erence between the left boundary (1. 3. 4) and the right boundary (1. 3. 5). The coe? cient H in Newton’s law of cooling is experimentally determined. It depends on properties of the rod as well as ? uid properties (including the ? uid velocity). If the coe? cient is very small, then very little heat energy ? ows across the boundary. In the limit as H > 0, Newton’s law of cooling approaches the insulated boundary condition. We can think of Newton’s law of cooling for H = 0 as representing an imperfectly insulated boundary.

If H > ? , the boundary condition approaches the one for prescribed temperature, u(0, t) = uB (t). This is most easily seen by dividing (1. 3. 4), for example, by H: ? K0 (0) ? u (0, t) = ? [u(0, t) ? uB (t)]. H ? x Thus, H > ? corresponds to no insulation at all. Summary. We have described three di? erent kinds of boundary conditions. For example, at x = 0, u(0, t) ? K0 (0) ? u (0, t) ? x ? K0 (0) ? u (0, t) ? x 3 For = = = uB (t) ? (t) ? H[u(0, t) ? uB (t)] prescribed temperature prescribed heat ? ux Newton’s law of cooling another situation in which (1. 3. 4) is valid, see Berg and McGregor [1966]. 4 Chapter 1. Heat Equation These same conditions could hold at x = L, noting that the change of sign (? H becoming H) is necessary for Newton’s law of cooling. One boundary condition occurs at each boundary. It is not necessary that both boundaries satisfy the same kind of boundary condition. For example, it is possible for x = 0 to have a prescribed oscillating temperature u(0, t) = 100 ? 25 cos t, and for the right end, x = L, to be insulated, ? u (L, t) = 0. ?x EXERCISES 1. 3 1. 3. 1. Consider a one-dimensional rod, 0 ? x ? L. Assume that the heat energy ? wing out of the rod at x = L is proportional to the temperature di? erence between the end temperature of the bar and the known external temperature. Derive (1. 3. 5) (brie? y, physically explain why H > 0). Two one-dimensional rods of di? erent materials joined at x = x0 are said to be in perfect thermal contact if the temperature is continuous at x = x0 : u(x0 ? , t) = u(x0 +, t) and no heat energy is lost at x = x0 (i. e. , the heat energy ? owing out of one ? ows into the other).

What mathematical equation represents the latter condition at x = x0 ? Under what special condition is ? u/? continuous at x = x0 ? *1. 3. 3. Consider a bath containing a ? uid of speci? c heat cf and mass density ? f that surrounds the end x = L of a one-dimensional rod. Suppose that the bath is rapidly stirred in a manner such that the bath temperature is approximately uniform throughout, equaling the temperature at x = L, u(L, t). Assume that the bath is thermally insulated except at its perfect thermal contact with the rod, where the bath may be heated or cooled by the rod. Determine an equation for the temperature in the bath. (This will be a boundary condition at the end x = L. ) (Hint: See Exercise 1. . 2. ) *1. 3. 2. 1. 4 1. 4. 1 Equilibrium Temperature Distribution Prescribed Temperature Let us now formulate a simple, but typical, problem of heat ? ow. If the thermal coe? cients are constant and there are no sources of thermal energy, then the temperature u(x, t) in a one-dimensional rod 0 ? x ? L satis? es ? 2u ? u = k 2. ?t ? x (1. 4. 1) 1. 4. Equilibrium Temperature Distribution 15 The solution of this partial di? erential equation must satisfy the initial condition u(x, 0) = f (x) (1. 4. 2) and one boundary condition at each end. For example, each end might be in contact with di? rent large baths, such that the temperature at each end is prescribed: u(0, t) u(L, t) = T1 (t) = T2 (t). (1. 4. 3) One aim of this text is to enable the reader to solve the problem speci? ed by (1. 4. 1– 1. 4. 3). Equilibrium temperature distribution. Before we begin to attack such an initial and boundary value problem for partial di? erential equations, we discuss a physically related question for ordinary di? erential equations. Suppose that the boundary conditions at x = 0 and x = L were steady (i. e. , independent of time), and u(L, t) = T2 , u(0, t) = T1 where T1 and T2 are given constants.

We de? ne an equilibrium or steady-state solution to be a temperature distribution that does not depend on time, that is, u(x, t) = u(x). Since ? /? t u(x) = 0, the partial di? erential equation becomes k(? 2 u/? x2 ) = 0, but partial derivatives are not necessary, and thus d2 u = 0. dx2 The boundary conditions are u(0) = T1 u(L) = T2 . (1. 4. 5) (1. 4. 4) In doing steady-state calculations, the initial conditions are usually ignored. Equation (1. 4. 4) is a rather trivial second-order ordinary di? erential equation (ODE). Its general solution may be obtained by integrating twice. Integrating (1. 4. ) yields du/dx = C1 , and integrating a second time shows that u(x) = C1 x + C2 . (1. 4. 6) We recognize (1. 4. 6) as the general equation of a straight line. Thus, from the boundary conditions (1. 4. 5) the equilibrium temperature distribution is the straight line that equals T1 at x = 0 and T2 at x = L, as sketched in Fig. 1. 4. 1. Geometrically there is a unique equilibrium solution for this problem. Algebraically, we 16 Chapter 1. Heat Equation can determine the two arbitrary constants, C1 and C2 , by applying the boundary conditions, u(0) = T1 and u(L) = T2 : u(0) = T1 u(L) = T2 implies implies T1 = C2 T2 = C1 L + C2 . 1. 4. 7) It is easy to solve (1. 4. 7) for the constants C2 = T1 and C1 = (T2 ? T1 )/L. Thus, the unique equilibrium solution for the steady-state heat equation with these ? xed boundary conditions is T2 ? T1 x. L (1. 4. 8) u(x) = T1 + u(x) T2 T1 x=0 x=L Figure 1. 4. 1 Equilibrium temperature distribution. Approach to equilibrium. For the time-dependent problem, (1. 4. 1) and (1. 4. 2), with steady boundary conditions (1. 4. 5), we expect the temperature distribution u(x, t) to change in time; it will not remain equal to its initial distribution f (x).

If we wait a very, very long time, we would imagine that the in? uence of the two ends should dominate. The initial conditions are usually forgotten. Eventually, the temperature is physically expected to approach the equilibrium temperature distribution, since the boundary conditions are independent of time: t>? lim u(x, t) = u(x) = T1 + T2 ? T1 x. L (1. 4. 9) In Sec. 8. 2 we will solve the time-dependent problem and show that (1. 4. 9) is satis? ed. However, if a steady state is approached, it is more easily obtained by directly solving the equilibrium problem. 1. 4. 2 Insulated Boundaries

As a second example of a steady-state calculation, we consider a one-dimensional rod again with no sources and with constant thermal properties, but this time with insulated boundaries at x = 0 and x = L. The formulation of the time-dependent 1. 4. Equilibrium Temperature Distribution problem is PDE: IC: BC1: BC2: ? u ? 2u =k 2 ? t ? x u(x, 0) = f (x) ? u (0, t) = 0 ? x ? u (L, t) = 0. ?x 17 (1. 4. 10) (1. 4. 11) (1. 4. 12) (1. 4. 13) The equilibrium problem is derived by setting ? u/? t = 0. The equilibrium temperature distribution satis? es d2 u =0 dx2 ODE : (1. 4. 14) BC1 : u (0) = 0 dx (1. 4. 15) BC2 : du (L) = 0, dx (1. 4. 16) where the initial condition is neglected (for the moment). The general solution of d2 u/dx2 = 0 is again an arbitrary straight line, u = C1 x + C2 . (1. 4. 17) The boundary conditions imply that the slope must be zero at both ends. Geometrically, any straight line that is ? at (zero slope) will satisfy (1. 4. 15) and (1. 4. 16), as illustrated in Fig. 1. 4. 2. u(x) Figure 1. 4. 2 Various constant equilibrium temperature distributions (with insulated ends). x=0 x=L The solution is any constant temperature. Algebraically, from (1. 4. 7), du/dx = C1 and both boundary conditions imply C1 = 0. Thus, u(x) = C2 (1. 4. 18) 18 Chapter

1. Heat Equation for any constant C2 . Unlike the ? rst example (with ? xed temperatures at both ends), here there is not a unique equilibrium temperature. Any constant temperature is an equilibrium temperature distribution for insulated boundary conditions. Thus, for the time-dependent initial value problem, we expect t>? lim u(x, t) = C2 ; if we wait long enough, a rod with insulated ends should approach a constant temperature. This seems physically quite reasonable. However, it does not ake sense that the solution should approach an arbitrary constant; we ought to know what constant it approaches. In this case, the lack of uniqueness was caused by the complete neglect of the initial condition. In general, the equilibrium solution will not satisfy the initial condition. However, the particular constant equilibrium solution is determined by considering the initial condition for the time-dependent problem (1. 4. 11). Since both ends are insulated, the total thermal energy is constant. This follows from the integral conservation of thermal energy of the entire rod [see (1. . 4)]: d dt L c? u dx = ? K0 0 ?u ? u (0, t) + K0 (L, t). ?x ? x (1. 4. 19) Since both ends are insulated, L c? u dx = constant. 0 (1. 4. 20) One implication of (1. 4. 20) is that the initial thermal energy must equal the ? L nal (limt>? ) thermal energy. The initial thermal energy is c? 0 f (x) dx since L u(x, 0) = f (x), while the equilibrium thermal energy is c? 0 C2 dx = c? C2 L since the equilibrium temperature distribution is a constant u(x, t) = C2 . The constant C2 is determined by equating these two expressions for the constant total therL mal energy, c? 0 f (x) dx = c?

C2 L. Solving for C2 shows that the desired unique steady-state solution should be 1 L L u(x) = C2 = f (x) dx, 0 (1. 4. 21) the average of the initial temperature distribution. It is as though the initial condition is not entirely forgotten. Later we will ? nd a u(x, t) that satis? es (1. 4. 10– 1. 4. 13) and show that limt>? u(x, t) is given by (1. 4. 21). EXERCISES 1. 4 1. 4. 1. Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions: 1. 4. Equilibrium Temperature Distribution ? a) Q = 0, (b) Q = 0, (c) Q = 0, ? (d) Q = 0, Q (e) = 1, K0 ? (f) Q = x2 , K0 u(0) = 0, u(0) = T , ? u (0) = 0, ? x u(0) = T , u(0) = T1 , u(0) = T , u(0) = T , u(L) = T u(L) = 0 u(L) = T ? u (L) = ? ?x u(L) = T2 ? u (L) = 0 ? x ? u (L) + u(L) = 0 ? x 19 (g) Q = 0, ? (h) Q = 0, ?u ? u (0) ? [u(0) ? T ] = 0, (L) = ? ?x ? x In these you may assume that u(x, 0) = f (x). 1. 4. 2. Consider the equilibrium temperature distribution for a uniform one-dimensional rod with sources Q/K0 = x of thermal energy, subject to the boundary conditions u(0) = 0 and u(L) = 0. (a) Determine the heat energy generated per unit time inside the entire rod. (b) Determine the heat energy ? owing out of the rod per unit time at x = 0 and at x = L. (c) What relationships should exist between the answers in parts (a) and (b)? 1. 4. 3. Determine the equilibrium temperature distribution for a one-dimensional rod composed of two di? erent materials in perfect thermal contact at x = 1. For 0 < x < 1, there is one material (c? = 1, K0 = 1) with a constant source (Q = 1), whereas for the other 1 < x < 2 there are no sources (Q = 0, c? = 2, K0 = 2) (see Exercise 1. 3. 2) with u(0) = 0 and u(2) = 0.

If both ends of a rod are insulated, derive from the partial di? erential equation that the total thermal energy in the rod is constant. Consider a one-dimensional rod 0 ? x ? L of known length and known constant thermal properties without sources. Suppose that the temperature is an unknown constant T at x = L. Determine T if we know (in the steady state) both the temperature and the heat ? ow at x = 0. The two ends of a uniform rod of length L are insulated. There is a constant source of thermal energy Q0 = 0, and the temperature is initially u(x, 0) = f (x). 1. 4. 4. 1. 4. 5. 1. 4. 6. 20 Chapter 1.

Heat Equation (a) Show mathematically that there does not exist any equilibrium temperature distribution. Brie? y explain physically. (b) Calculate the total thermal energy in the entire rod. 1. 4. 7. For the following problems, determine an equilibrium temperature distribution (if one exists). For what values of ? are there solutions? Explain physically. ? (a) ? u ? 2u = + 1, ? t ? x2 ? 2u ? u = , ? t ? x2 u(x, 0) = f (x), ? u (0, t) = 1, ? x ? u (0, t) = 1, ? x ? u (0, t) = 0, ? x ? u (L, t) = ? ?x ? u (L, t) = ? ?x ? u (L, t) = 0 ? x (b) u(x, 0) = f (x), (c) 1. 4. 8. ?2u ? u = + x ? ?, u(x, 0) = f (x), ? ? x2 Express the integral conservation law for the entire rod with constant thermal properties. Assume the heat ? ow is known to be di? erent constants at both ends By integrating with respect to time, determine the total thermal energy in the rod. (Hint: use the initial condition. ) (a) Assume there are no sources. (b) Assume the sources of thermal energy are constant. 1. 4. 9. Derive the integral conservation law for the entire rod with constant thermal properties by integrating the heat equation (1. 2. 10) (assuming no sources). Show the result is equivalent to (1. 2. 4). 2 1. 4. 10. Suppose ? u = ? + 4, u(x, 0) = f (x), ? u (0, t) = 5, ? u (L, t) = 6. Calculate ? t ? x2 ? x ? x the total thermal energy in the one-dimensional rod (as a function of time). 1. 4. 11. Suppose ? u ? t = ?2u ? x2 + x, u(x, 0) = f (x), ?u ? x (0, t) = ? , ?u ? x (L, t) = 7. (a) Calculate the total thermal energy in the one-dimensional rod (as a function of time). (b) From part (a), determine a value of ? for which an equilibrium exists. For this value of ? , determine lim u(x, t). t>? 1. 4. 12. Suppose the concentration u(x, t) of a chemical satis? es Fick’s law (1. 2. 13), and the initial concentration is given u(x, 0) = f (x).

Consider a region 0 < x < L in which the ? ow is speci? ed at both ends ? k ? u (0, t) = ? and ? x ? k ? u (L, t) = ?. Assume ? and ? are constants. ?x (a) Express the conservation law for the entire region. (b) Determine the total amount of chemical in the region as a function of time (using the initial condition). 1. 5. Heat Equation in Two or Three Dimensions 21 (c) Under what conditions is there an equilibrium chemical concentration and what is it? 1. 4. 13. Do Exercise 1. 4. 12 if ? and ? are functions of time. 1. 5 Derivation of the Heat Equation in Two or Three Dimensions Introduction. In Sec. 1. we showed that for the conduction of heat in a onedimensional rod the temperature u(x, t) satis? es c? ? ? u = ? t ? x K0 ? u ? x + Q. In cases in which there are no sources (Q = 0) and the thermal properties are constant, the partial di? erential equation becomes ? 2u ? u = k 2, ? t ? x where k = K0 /c?. Before we solve problems involving these partial di? erential equations, we will formulate partial di? erential equations corresponding to heat ? ow problems in two or three spatial dimensions. We will ? nd the derivation to be similar to the one used for one-dimensional problems, although important di? erences will emerge.

We propose to derive new and more complex equations (before solving the simpler ones) so that, when we do discuss techniques for the solutions of PDEs, we will have more than one example to work with. Heat energy. We begin our derivation by considering any arbitrary subregion R, as illustrated in Fig. 1. 5. 1. As in the one-dimensional case, conservation of heat energy is summarized by the following word equation: heat energy ? owing rate of change heat energy generated = across the boundaries + of heat energy inside per unit time, per unit time where the heat energy within an arbitrary subregion R is heat energy = R ? u dV, instead of the one-dimensional integral used in Sec. 1. 2. 22 Chapter 1. Heat Equation Figure 1. 5. 1 Three-dimensional subregion R. ? ?? n ? ? ? A B Figure 1. 5. 2 Outward normal component of heat ? ux vector. Heat ? ux vector and normal vectors. We need an expression for the ? ow of heat energy. In a one-dimensional problem the heat ? ux ? is de? ned to the right (? < 0 means ? owing to the left). In a three-dimensional problem the heat ? ows in some direction, and hence the heat ? ux is a vector ?. The magnitude of ? is the amount of heat energy ? owing per unit time per unit surface area.

However, in considering conservation of heat energy, it is only the heat ? owing across the boundaries per unit time that is important. If, as at point A in Fig. 1. 5. 2, the heat ? ow is parallel to the boundary, then there is no heat energy crossing the boundary at that point. In fact, it is only the normal component of the heat ? ow that contributes (as illustrated by point B in Fig. 1. 5. 2). At any point there are two normal vectors, an inward and an outward normal n. We will use the convention of only utilizing the unit outward normal vector n (where the ? stands for a ? unit vector). Conservation of heat energy.

At each point the amount of heat energy ? owing out of the region R per unit time per unit surface area is the outward normal component of the heat ? ux vector. From Fig. 1. 5. 2 at point B, the outward normal component of the heat ? ux vector is |? | cos ? = ? · n/|n| = ? · n . If the heat ? ux ? vector ? is directed inward, then ? · n < 0 and the outward ? ow of heat energy ? is negative. To calculate the total heat energy ? owing out of R per unit time, we must multiply ? · n by the di? erential surface area dS and “sum” over the entire ? surface that encloses the region R. This4 is indicated by the closed surface integral ? · n dS. This is the amount of heat energy (per unit time) leaving the region ? ?? R and (if positive) results in a decreasing of the total heat energy within R. If Q 4 Sometimes the notation ? is used instead of ? · n, meaning the outward normal component ? n of ?. 1. 5. Heat Equation in Two or Three Dimensions 23 is the rate of heat energy generated per unit volume, then the total heat energy Q dV . Consequently, conservation of heat energy generated per unit time is R for an arbitrary three-dimensional region R becomes d dt R ? ? c? u dV = ? ? ?? · n dS + R Q dV. (1. 5. 1) Divergence theorem.

In one dimension, a way in which we derived a partial di? erential relationship from the integral conservation law was to notice (via the fundamental theorem of calculus) that b ?(a) ? ?(b) = ? a ?? dx; ? x that is, the ? ow through the boundaries can be expressed as an integral over the entire region for one-dimensional problems. We claim that the divergence theorem is an analogous procedure for functions of three variables. The divergence theorem ? i+Ay ? +Az k) j deals with a vector A (with components Ax , Ay and Az ; i. e. , A = Ax? and its divergence de? ned as follows: ? ·A ? ? ? ? Ax + Ay + Az . ?x ? ? z (1. 5. 2) Note that the divergence of a vector is a scalar. The divergence theorem states that the volume integral of the divergence of any continuously di? erentiable vector A is the closed surface integral of the outward normal component of A: ? ?·A dV = ? A · n dS. ?? R (1. 5. 3) This is also known as Gauss’s theorem. It can be used to relate certain surface integrals to volume integrals, and vice versa. It is very important and very useful (both immediately and later in this text). We omit a derivation, which may be based on repeating the one-dimensional fundamental theorem in all three dimensions.

Application of the divergence theorem to heat ? ow. In particular, the closed surface integral that arises in the conservation of heat energy (1. 5. 1), corresponding to the heat energy ? owing across the boundary per unit time, can be written as a volume integral according to the divergence theorem, (1. 5. 3). Thus, (1. 5. 1) becomes d dt R c? u dV = ? R ?·? dV + R Q dV. (1. 5. 4) 24 Chapter 1. Heat Equation We note that the time derivative in (1. 5. 4) can be put inside the integral (since R is ? xed in space) if the time derivative is changed to a partial derivative. Thus, all the expressions in (1. 5. ) are volume integrals over the same volume, and they can be combined into one integral: c? R ?u + ? ·? ? Q ? t dV = 0. (1. 5. 5) Since this integral is zero for all regions R, it follows (as it did for one-dimensional integrals) that the integrand itself must be zero: c? or, equivalently, c? ?u = ?? ·? + Q. ?t (1. 5. 6) ? u + ? ·? ? Q = 0 ? t Equation (1. 5. 6) reduces to (1. 2. 3) in the one-dimensional case. Fourier’s law of heat conduction. In one-dimensional problems, from experiments according to Fourier’s law, the heat ? ux ? is proportional to the derivative of the temperature, ? = ? K0 ? u/? x.

The minus sign is related to the fact that thermal energy ? ows from hot to cold. ?u/? x is the change in temperature per unit length. These same ideas are valid in three dimensions. In the appendix, we derive that the heat ? ux vector ? is proportional to the temperature gradient ? ?u ? ?u ? + ? u ? + ? u k : ? x i ? y j ? z ? = ? K0 ? u, (1. 5. 7) known as Fourier’s law of heat conduction, where again K0 is called the thermal conductivity. Thus, in three dimensions the gradient ? u replaces ? u/? x. Heat equation. When the heat ? ux vector, (1. 5. 7), is substituted into the conservation of heat energy equation, (1. . 6), a partial di? erential equation for the temperature results: c? ?u = ? ·(K0 ? u) + Q. ?t (1. 5. 8) In the cases in which there are no sources of heat energy (Q = 0) and the thermal coe? cients are constant, (1. 5. 8) becomes ? u = k? ·(? u), ? t (1. 5. 9) 1. 5. Heat Equation in Two or Three Dimensions 25 where k = K0 /c? is again called the thermal di? usivity. From their de? nitions, we calculate the divergence of the gradient of u: ? ?x ? u ? x ? ?y ? u ? y ? ?z ? u ? z ? 2u ? 2u ? 2u + 2 + 2 ? ?2 u. ?x2 ? y ? z ?·(? u) = + + = (1. 5. 10) This expression ? 2 u is de? ned to be the Laplacian of u.

Thus, in this case ? u = k? 2 u. ?t (1. 5. 11) Equation (1. 5. 11) is often known as the heat or di? usion equation in three spatial dimensions. The notation ? 2 u is often used to emphasize the role of the del operator ? : ? ? ? ? ?? i+ j+ k. ?? ?x ? y ? z Note that ? u is ? operating on u, while ? ·A is the vector dot product of del with A. Furthermore, ? 2 is the dot product of the del operator with itself or ? ·? = ? ?x ? ?x + ? ?y ? ?y + ? ?z ? ?z operating on u, hence the notation del squared, ? 2 . Initial boundary value problem. In addition to (1. 5. 8) or (1. 5. 11), the temperature satis? s a given initial distribution, u(x, y, z, 0) = f (x, y, z). The temperature also satis? es a boundary condition at every point on the surface that encloses the region of interest. The boundary condition can be of various types (as in the one-dimensional problem). The temperature could be prescribed, u(x, y, z, t) = T (x, y, z, t), everywhere on the boundary where T is a known function of t at each point of the boundary. It is also possible that the ? ow across the boundary is prescribed. Frequently, we might have the boundary (or part of the boundary) insulated. This means that there is no heat ? w across that portion of the boundary. Since the heat ? ux vector is ? K0 ? u, the heat ? owing out will be the unit outward normal component of the heat ? ow vector, ? K0 ? u · n, where n is a unit outward normal ? ? to the boundary surface.

Thus, at an insulated surface, ? u·? = 0. n 26 Chapter 1. Heat Equation Recall that ? u·? is the directional derivative of u in the outward normal direction; n it is also called the normal derivative. 5 Often Newton’s law of cooling is a more realistic condition at the boundary. It states that the heat energy ? owing out per unit time per unit surface area is proportional to the di? rence between the temperature at the surface u and the temperature outside the surface ub . Thus, if Newton’s law of cooling is valid, then at the boundary n ? K0 ? u·? = H(u ? ub ). (1. 5. 12) Note that usually the proportionality constant H > 0, since if u > ub , then we exn pect that heat energy will ? ow out and ? K0 ? u·? will be greater than zero. Equation (1. 5. 12) veri? es the two forms of Newton’s law of cooling for one-dimensional problems. In particular, at x = 0, n = ?? and the left-hand side (l. h. s. ) of (1. 5. 12) ? i ? i becomes K0 ? u/? x, while at x = L, n = ? and the l. h. s. of (1. . 12) becomes ? K0 ? u/? x [see (1. 3. 4) and (1. 3. 5)]. Steady state. If the boundary conditions and any sources of thermal energy are independent of time, it is possible that there exist steady-state solutions to the heat equation satisfying the given steady boundary condition: 0 = ? ·(K0 ? u) + Q. Note that an equilibrium temperature distribution u(x, y, z) satis? es a partial di? erential equation when more than one spatial dimension is involved. In the case with constant thermal properties, the equilibrium temperature distribution will satisfy ? 2 u = ? Q , K0 (1. 5. 13) known as Poisson’s equation.

If, in addition, there are no sources (Q = 0), then ?2 u = 0; (1. 5. 14) the Laplacian of the temperature distribution is zero. Equation (1. 5. 14) is known as Laplace’s equation. It is also known as the potential equation, since the gravitational and electrostatic potentials satisfy (1. 5. 14) if there are no sources. We will solve a number of problems involving Laplace’s equation in later sections. 5 Sometimes (in other books and references) the notation ? u/? n is used. However, to calculate ? u/? n we usually calculate the dot product of the two vectors, ? u and n, ? u·? , so we will not ? n use the notation ? /? n in this text. 1. 5. Heat Equation in Two or Three Dimensions 27 Two-dimensional problems. All the previous remarks about threedimensional problems are valid if the geometry is such that the temperature only depends on x, y and t. For example, Laplace’s equation in two dimensions, x and y, corresponding to equilibrium heat ? ow with no sources (and constant thermal properties) is ? 2 u = ? 2u ? 2u + 2 = 0, ? x2 ? y since ? 2 u/? z 2 = 0. Two-dimensional results can be derived directly (without taking a limit of three-dimensional problems), by using fundamental principles in two dimensions.

We will not repeat the derivation. However, we can easily outline the results. Every time a volume integral ( R · · · dV ) appears, it must be replaced by a surface integral over the entire two-dimensional plane region ( R · · · dS). Similarly, the boundary contribution for three-dimensional problems, which is the closed ? surface integral ? ? · · · dS, must be replaced by the closed line integral · · · d? , an integration over the boundary of the two-dimensional plane surface. These results are not di? cult to derive since the divergence theorem in three dimensions, ? ?·A dV = ? ?A · n dS, ? R 1. 5. 15) is valid in two dimensions, taking the form ? ·A dS = R A · n d?. ? (1. 5. 16) Sometimes (1. 5. 16) is called Green’s theorem, but we prefer to refer to it as the twodimensional divergence theorem. In this way only one equation need be familiar to the reader, namely (1. 5. 15); the conversion to two-dimensional form involves only changing the number of integral signs. Polar and cylindrical coordinates. The Laplacian, ?2 u = ? 2u ? 2u ? 2u + 2 + 2, ? x2 ? y ? z (1. 5. 17) is important for the heat equation (1. 5. 11) and its steady-state version (1. 5. 14), as well as for other signi? ant problems in science and engineering. Equation (1. 5. 17) written in (1. 5. 17) in Cartesian coordinates is most useful when the geometrical region under investigation is a rectangle or a rectangular box. Other coordinate systems are frequently useful. In practical applications, we may need the formula that expresses the Laplacian in the appropriate coordinate system. In circular cylindrical coordinates, with r the radial distance from the z-axis and ? the angle 28 Chapter 1. Heat Equation x = r cos ? y = r sin ? z = z, the Laplacian can be shown to equal the following formula: 1 ? r ? r ? u ? 1 ? 2u ? 2u + 2. r2 ?? 2 ? z (1. 5. 18) ?2 u = r + (1. 5. 19) There may be no need to memorize this formula, as it can often be looked up in a reference book. As an aid in minimizing errors, it should be noted that every term in the Laplacian has the dimension of u divided by two spatial dimensions [just as in Cartesian coordinates, (1. 5. 17)]. Since ? is measured in radians, which have no dimensions, this remark aids in remembering to divide ? 2 u/?? 2 by r2 . In polar coordinates (by which we mean a two-dimensional coordinate system with z ? xed, usually z = 0), the Laplacian is the same as (1. . 19) with ? 2 u/? z 2 = 0 since there is no dependence on z. Equation (1. 5. 19) can be derived (see the Exercises) using the chain rule for partial derivatives, applicable for changes of variables. In some physical situations it is known that the temperature does not depend on the polar angle ? ; it is said to be circularly or axially symmetric. In that case ? 2 u = 1 ? r ? r r ? u ? r + ? 2u . ?z 2 (1. 5. 20) Spherical coordinates. Geophysical problems as well as electrical problems with spherical conductors are best solved using spherical coordinates (? , ? , ? ).

The radial distance is ? , the angle from the pole (z-axis) is ? , and the cylindrical (or azimuthal) angle is ?. Note that if ? is constant and the angle ? is a constant a circle is generated with radius ? sin ? (as shown in Fig. 1. 5. 3) so that x = ? sin ? cos ? y = ? sin ? sin ? z = ? cos ?. (1. 5. 21) The angle from the pole ranges from 0 to ? (while the usual cylindrical angle ranges from 0 to 2? ). It can be shown that the Laplacian satis? es 1 ? ?2 ?? ?u ?? ? 1 2 sin ? ?? ? ? u ?? ?2u 1 . ?2 sin2 ? ?? 2 ?2 u = ?2 + sin ? + (1. 5. 22) 1. 5. Heat Equation in Two or Three Dimensions =L 29 Area magnified ? ? ? sin? ? Figure 1. 5. 3 Spherical coordinates. EXERCISES 1. 5 1. 5. 1. Let c(x, y, z, t) denote the concentration of a pollutant (the amount per unit volume). (a) What is an expression for the total amount of pollutant in the region R? (b) Suppose that the ? ow J of the pollutant is proportional to the gradient of the concentration. (Is this reasonable? ) Express conservation of the pollutant. (c) Derive the partial di? erential equation governing the di? usion of the pollutant. *1. 5. 2. For conduction of thermal energy, the heat ? ux vector is ? = ? K0 ? u.

If in addition the molecules move at an average velocity V , a process called convection, then brie? y explain why ? = ? K0 ? u + c? uV . Derive the corresponding equation for heat ? ow, including both conduction and convection of thermal energy (assuming constant thermal properties with no sources). Consider the polar coordinates x = r cos ? y = r sin ?. ?r ? r (a) Since r2 = x2 + y 2 , show that ? x = cos ? , ? y = sin ? , cos ? ?? ? sin ? . r , and ? x = r ? (b) Show that r = cos ?? + sin ?? and ? = ? sin ?? + cos ?? ? i j i j. ?? ?y 1. 5. 3. = ? ? (c) Using the chain rule, show that ? = r ? r + ? 1 ?? nd hence ? u = ?? r ? u ? 1 ? u ? ?r r + r ?? ?. ? ? ? (d) If A = Ar r + A? ?, show that ? ·A = 1 ? r (rAr ) + 1 ?? (A? ), since ? r r ? ? ? r /?? = ? and ? ?/?? = ?? follows from part (b). ? r 30 (e) Show that ? 2 u = 1. 5. 4. 1. 5. 5. 1 ? r ? r Chapter 1. Heat Equation r ? u + ? r 1 ? 2u r 2 ?? 2 . Using Exercise 1. 5. 3(a) and the chain rule for partial derivatives, derive the special case of Exercise 1. 5. 3(e) if u(r) only. Assume that the temperature is circularly symmetric: u = u(r, t), where r2 = x2 + y 2 . We will derive the heat equation for this problem. Consider any circular annulus a ? r ? . (a) Show that the total heat energy is 2? b a c? ur dr. (b) Show that the ? ow of heat energy per unit time out of the annulus at r = b is ? 2? bK0 ? u/? r |r=b . A similar result holds at r = a. (c) Use parts (a) and (b) to derive the circularly symmetric heat equation without sources: k ? ?u ? u = r . ?t r ? r ? r 1. 5. 6. 1. 5. 7. 1. 5. 8. Modify Exercise 1. 5. 5 if the thermal properties depend on r. Derive the heat equation in two dimensions by using Green’s theorem, (1. 5. 16), the two-dimensional form of the divergence theorem. If Laplace’s equation is satis? ed in three dimensions, show that ? ? ?? u·? dS = 0 for any closed surface. (Hint: Use the divergence theorem. ) Give a physical interpretation of this result (in the context of heat ? ow). 1. 5. 9. Determine the equilibrium temperature distribution inside a circular annulus (r1 ? r ? r2 ): *(a) if the outer radius is at temperature T2 and the inner at T1 (b) if the outer radius is insulated and the inner radius is at temperature T1 1. 5. 10. Determine the equilibrium temperature distribution inside a circle (r ? r0 ) if the boundary is ? xed at temperature T0 . *1. 5. 11. Consider k ? ?u = ? t r ? r ? u ? r r aa